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Question
Show that the point (11, – 2) is equidistant from (4, – 3) and (6, 3)
Solution
Let P(x1, y1) = P(11, – 2), Q(x2, y2) = Q(4, – 3), R(x3, y3) = R(6, 3)
By distance formula,
d(P, Q) = `sqrt((x_2 - x_1)^2 + (y_2 - y_1)^2`
= `sqrt((4 - 11)^2 + [-3 - (-2)]^2`
= `sqrt((-7)^2 + (-1)^2`
= `sqrt(49 + 1)`
= `sqrt(50)`
= `5sqrt(2)`
And
d(P, R) = `sqrt((x_3 - x_1)^2 + (y_3 - y_1)^2`
= `sqrt((6 - 11)^2 + [3 - (-2)]^2`
= `sqrt((-5)^2 + (5)^2`
= `sqrt(25 + 25)`
= `sqrt(50)`
= `5sqrt(2)`
Here, d(P, Q) = d(P, R)
∴ Point (11, – 2) is equidistant from (4, – 3) and (6, 3).
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